Question

Let $f(x) = \frac{x^2}{x^2 - 1}.$find the largest integer $n$ so that $f(2) \cdot f(3) \cdot f(4) \cdots f(n-1) \cdot f(n) < 1.98.$

Answer 1

The question asks: "Let $f(x) = \frac{ x^{2} }{ x^{2} - 1}$. Find the largest integer n so that f(2) · f(3) · f(4) · ... · f(n-1) · f(n) < 1.98"

The answer is n = 98

Explanation:

First thing, consider that the function can be written as:
$\frac{ x^{2} }{ x^{2} - 1} = \frac{x \cdot x}{(x - 1)(x + 1)}$

Now, let's expand the product, substituting the function with its equation for the requested values:
$\frac{2 \cdot 2}{(1)(3)} \cdot \frac{3 \cdot 3}{(2)(4)} \cdot \frac{4 \cdot 4}{(3)(5)} \cdot ... \cdot \frac{(n-1) \cdot (n-1)}{(n-2))(n)} \cdot \frac{n \cdot n}{(n - 1)(n + 1)} \ \textless \ 1.98$

As you can see, the intermediate terms cancel out with each other, leaving us with:
$\frac{2 \cdot n}{1 \cdot (n+1)} \ \textless \ 1.98$

This is a simple inequality that can be easily solved:
$\frac{2n}{n+1} \ \textless \frac{198}{100}$

200n < 198(n + 1)
200n < 198n + 198
2n < 198
n < 99

Hence, the greatest integer n < 99 (extremity excluded) is 98.